Variable vs constant power supply

[Tom.G]

In your description, did you mean the voltage is always 12V even when current vary from 0 to 5A?

Yes!

when the current drawn is less than 5A, the voltage will be greater like 18V (for no load). It is only when the current drawn is 5A that the voltage will be 12V at full load. In both case the power in watt is maintained from voltage times ampere).

Sorry, not quite correct, the delivered power is not constant. Please see the datasheet below for a table of how the output voltage of an typical unregulated supply varies as the current changes.
Beware of the graph on that page, the x-axis is nonlinear. At low currents the voltage drops rapidly with increasing loads.

You will see that at 50mA load the delivered power is 1Watt;
but at 300mA load the delivered power is 4.8W!

https://assets.omega.com/manuals/M1388.pdf

Cheers,
Tom

 

[Ephant]

Yes!Sorry, not quite correct, the delivered power is not constant. Please see the datasheet below for a table of how the output voltage of an typical unregulated supply varies as the current changes.
Beware of the graph on that page, the x-axis is nonlinear. At low currents the voltage drops rapidly with increasing loads.

You will see that at 50mA load the delivered power is 1Watt;
but at 300mA load the delivered power is 4.8W!

https://assets.omega.com/manuals/M1388.pdf

Cheers,
Tom

How do you know what wattage is the right one, 1Watt for 50mA load or 4.8W for 300mA load? If both is wrong, how do you check the right wattage?

 

[Tom.G]

how do you check the right wattage?

"...right wattage?"

That is sort of like asking "What is the weight you can lift?". There is no "right" answer because you can lift your limit or anything below your limit.

The Maximum wattage that the power pack can supply is stated, sometimes indirectly, on the pack itself or in the data sheet.

If not explicitly stated, the Wattage is the normal output voltage, V, times the rated output current I; in other words W = V x I -- where V is the output voltage, and I is the current that the load is using (drawing from the power pack).